Check if $\ln(x), x > 0$ is uniformly continuous 
Check if $\ln(x), x > 0$ is uniformly continuous

My only idea on solving this was to use the definition of uniform continuity. Namely, I need to show that for all $\epsilon >0$ there exists a $\delta >0$ such that for all $x_1, x_2$ in the domain of the function $|x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$
I have never before proved anything like this and I have no idea whether or not this function is uniformly continuous. Anyway, looking at the tremendous slope close to zero I am inclined to believe that this function is not uniformly continuous. 

I want to show that
$$(\exists \epsilon>0)(\forall \delta>0)(\exists x_1, x_2)(|f(x_1)-f(x_2)| \ge \epsilon \Rightarrow |x_1-x_2| \ge \delta$$
And so let's pick $\epsilon = 1$
I want to show that we can pick such $x_1, x_2$ that $|\ln(x_1)- \ln(x_2)| > 1 \Rightarrow |x_1 - x_2| > \delta$ for all $\delta$. 
We know that $x_1 \ge ex_2$. 
And now, let's take an arbitrary $\delta$, such that 
$$|x_1 - x_2| \ge \delta$$
Now, we can simply pick
$$x_1 = ex_2 + \delta$$
And
$$x_2 = 1$$
And it works. And so the function is not uniformly continuous. 

I know that my solution is very chaotic and confusing,is there an easier way to tackle this?
 A: A good criterion (theorem) to check whether a function is uniformly continuous or not is the following theorem, which is specially useful and powerful when you want to disprove something is uniformly continuous:
Theorem: A function is uniformly continuous on $A$ if and only if for any two sequences $\{a_n\}$ and $\{b_n\}$ in $A$:
$$
\lim_{n\to\infty}\left(a_n-b_n\right)=0 \implies \lim_{n\to\infty}\left(f(a_n)-f(b_n)\right)=0.
$$
Now to disprove that $\ln(x)$ is uniformly continuous on $(0,+\infty)$, take $a_n=\{\frac{1}{n}\}$ and $b_n=\{\frac{1}{n^2}\}$
Then $$\lim_{n\to\infty}(a_n-b_n)=\lim_{n\to\infty}(\frac{1}{n}-\frac{1}{n^2})=\lim_{n\to\infty}\frac{n-1}{n^2}=0$$ while $\lim_{n\to\infty} \left(\ln(\frac{1}{n})-\ln(\frac{1}{n^2})\right)=\lim_{n\to\infty}\ln(\frac{n^2}{n})=\ln(+\infty)=+\infty$
Q.E.D.
Remarks:
Note that uniform continuity very much depends on the given set and therefore, it's generally very hard to establish just by going through the $\epsilon-\delta$ definition. For example, if we wanted to study whether $f$ is uniformly continuous or not on $[1,+\infty)$ the answer would've been positive. Indeed, $$|\frac{\mathrm{d}}{\mathrm{d}x}\ln(x)|=|\frac{1}{x}|\leq 1  \text{ for every x}\in [1,+\infty)$$
Therefore, $\ln(x)$ is uniformly continuous on $[1,+\infty)$.
Also, remember that every continuous function is uniformly continuous on a compact set.
A: 
I want to show that
  $$(\exists \epsilon>0)(\forall \delta>0)(\exists x_1, x_2)(|f(x_1)-f(x_2)| \ge \epsilon \Rightarrow |x_1-x_2| \ge \delta$$

This is not the correct negation of the definition of uniform continuity. It should instead read as follows.

I want to show that there exists $\epsilon > 0$ such that for all $\delta > 0$ there exist $x_1, x_2$ in the domain of $f$ satisfying $|x_1 - x_2| < \delta$ and $|f(x_1) - f(x_2)| \geq \epsilon$.

So, to show that $\ln(x)$ is not uniformly continuous, we can try $\epsilon = 1$ just as you have written. Now,
$$
\begin{align*}
&\ |\ln(x_1) - \ln(x_2)| \geq 1\\
\Leftrightarrow &\ |\ln(x_1/x_2)| \geq 1\\
\Leftrightarrow &\ \ln(x_1/x_2) \leq -1 \text{ or } \ln(x_1/x_2) \geq 1\\
\Leftrightarrow &\ x_1/x_2 \leq e^{-1} \text{ or } x_1/ x_2 \geq e
\end{align*}
$$
So, for each $\delta > 0$, we will search for points $x_1, x_2 > 0$ satisfying $x_2 = ex_1$, such that $|x_1 - x_2| < \delta$. But such $x_1, x_2$ can always be found, because we only need to choose $x_1 = \frac{\delta}{2(e-1)}$. Then, $$|x_1 - x_2| = |x_1 - e x_1| = |x_1| \cdot |1-e| < \frac{\delta}{2(e-1)}(e-1) = \delta.$$
Hence, $f(x) = \ln(x)$ is not uniformly continuous.
